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Old May 18th 2009, 09:46 PM   #1
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Refraction and Snell's Law

Part 1: The greatest recorded thickness of ice is 4.78 km. It was measured in Antarctica by using the reflection of radio waves. Suppose visible light could be used as well as radio waves. If the light enters the ice with an angle of incidence of 33 degrees, reflects from the bottom of the ice, and reemerges 4.4 km away, what is the index of refraction for the ice?

Part 2: What would the distance between the entry and exit points in item 2 be if the light entered the ice from a layer of water that lay over the ice?


Okay. I'll try to describe how I got my answers. Please correct me if I do something wrong. :P


Part 1:
I drew a picture and created a right triangle with leg measurements of 2200m for the horizontal leg and 4780m for the vertical leg.
Then, I found the angle measurement between the hypotenuse and the horizontal leg by finding the inverse tangent of 4780/2200, which I got to be about 65.2857 degrees.
I subtracted that from 90 degrees because I need the the angle of refraction (I think that's what it's called), between the triangle's hypotenuse and the normal (perpendicular created by the light beam intersecting with the ice), using the complementary thing.
Next, I used Snell's Law and the equation looks like this:
(1)(sin 33)=(n)(sin 24.7143)
which turns into
n=(sin 33)/(sin 24.7143)
and I got the index of refraction of ice to be 1.3027. I Googled the index of refraction of ice and it's 1.31. It SEEMS correct, but I just wanna make sure. :P


Part 2:
First, I find the angle of refraction so I use Snell's Law again:
(1.333)(sin 33)=(1.3027)(sin x)
and I get 33.8698 degrees
Then, I use the complementary thingy again (lol. idk.) and 90 - 33.8698 = 56.1302.
Using this angle measurement, I create a right triangle similar to the one in part 1.
And I set up an equation with the information from the constructed right triangle:
(tan 56.1302) = 4780/x
and x = 3208.3708
Then I double x and I get 6416.7417 as my answer.

Can somebody confirm or correct me?

(I posted this in the mathhelpforum then I saw the announcement thread thing about physicshelpforum. )
iEricKim is offline   Reply With Quote
Old May 19th 2009, 01:36 AM   #2
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Part 1 is fine.

In part 2 you should use

(1.0)(sin 33)=(refractive index of water)(sin x).
This angle x will be the angle of incidence for the water ice interface and repeat. If you check your diagram you will find that x and the new angle of incidence are equal alternate angles i think not 'complementary thingy '!
physicsquest is offline   Reply With Quote
Old May 19th 2009, 01:41 AM   #3
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Originally Posted by physicsquest View Post
equal alternate angles i think not 'complementary thingy '!
Lol.
Thanks.
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